A conjecture about lines and points in the plane

Question

Let $ l_1, l_2, \ldots$ be an infinite sequence of lines in the plane, and let $(a_1, b_1), (a_2, b_2), \ldots$ be an infinite sequence of pairs of points such that $a_i, b_i \in l_i$ and $a_i \neq b_i$, so that any pair $(a_i, b_i)$ defines an open interval of $l_i$.

Conjecture: it is always possible to choose a point $x_i$ from every interval $(a_i, b_i)$ such that there is no line in the plane containing countably many of these points.

Can somebody help me to prove or disprove this conjecture?

Answer 1

Let $x_1$ be any point in $(a_1, b_1) \backslash \bigcup \{l_i : i \neq 1\}$. Suppose $x_1, x_2, \dots, x_n$ have already been chosen such that $x_i \notin l_j$ for any $j \neq i$, no three $x_i$'s are collinear. Now choose $x_{n+1}$ as follows: List all lines $l$ that pass through two of the $x_i$'s and note that these lines can meet $l_{n+1}$ at at most one point. Choose $x_{n+1} \in (a_{n+1}, b_{n+1}) \backslash \bigcup \{l_j : j \neq i\}$ different from these points.

Answer 2

Let's say that the set of points has the property $(*)$ if any three of them don't lie on the same line. Now, let's show that we can choose such $x_i\in(a_i,b_i)$ that will have the property $(*)$.

Say, you already choose $x_1,\dots,x_n$ such that they have the property $(*)$. Now, in worst case, if for any $x\in(a_{n+1},b_{n+1})$ there is $x_i$ and $x_j$ such that $x,x_i,x_j$ lie on the same line, then fix any $x=x_{n+1}\in(a_{n+1},b_{n+1})$ and move the point, say, $x_i$ "slightly" on the interval $(a_i,b_i)$. First, by changing the position of the point $x_i$ will exclude that $x_{n+1},x_i,x_j$ lie on the same line. Secondly, since the initial points $x_1,\dots,x_n$ have the property $(*)$, then without breaking that property $(*)$ we can change the position of one point, at least a little.

Thus, we've extended our set of points $x_i$, protecting the property $(*)$.